Question #121873
Let f (x) = sin(ax + b) and let g (x) = cos(ax + b) where a and b are constants. Gues a formular for f*(x) and g"(x) for general n (positive integer )
1
Expert's answer
2020-06-16T19:16:03-0400

Given f(x)=sin(ax+b),g(x)=cos(ax+b)f (x) = sin(ax + b), g(x) = cos(ax+b)

    f(x)=a cos(ax+b),g(x)=a sin(ax+b)\implies f'(x) = a \ cos(ax+b), g'(x) = -a \ sin(ax+b)

    f(x)=a2sin(ax+b),g(x)=a2cos(ax+b)\implies f''(x) = -a^2 sin(ax+b), g''(x)=-a^2cos(ax+b)

    f(x)=a3cos(ax+b),g(x)=a3sin(ax+b)    fiv(x)=a4sin(ax+b),giv(x)=a4cos(ax+b)\implies f'''(x) = -a^3 cos(ax+b), g'''(x) = a^3 sin(ax+b) \\ \implies f^{iv}(x) = a^4 sin(ax+b), g^{iv}(x) = a^4 cos(ax+b)

Hence In general,

fn(x)={(1)n2ansin(ax+b):n is even(1)n12ancos(ax+b):n is oddf^n(x) = \begin{cases} (-1)^{\frac{n}{2}} a^n sin(ax+b) : n \ is \ even \\ (-1)^{\frac{n-1}{2}} a^n cos(ax+b) : n \ is \ odd \end{cases}

and gn(x)={(1)n2ancos(ax+b):n is even(1)n+12ansin(ax+b):n is oddg^n(x) = \begin{cases} (-1)^{\frac{n}{2}} a^n cos(ax+b) : n \ is \ even \\ (-1)^{\frac{n+1}{2}} a^n sin(ax+b) : n \ is \ odd \end{cases}


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